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Scheduling problem

  1. Jun 26, 2005 #1
    The probability that project a will be complete at time t after it begins is given as:

    [tex]
    \[ p_a(t) = \left\{ \begin{array}{ll}
    0 & \mbox{if t \leq t_1$} \\
    \frac {(t - t_1)}{t_2-t_1} & \mbox{if t_1 \leq t \leq t_2$} \\
    1 & \mbox{if t_2 \leq t}
    \right. \][/tex]

    In other words, the project requires at least [itex]t_1[/itex] to complete and will take no more than [itex]t_2[/itex]. The probability of completion at any time between [itex]t_1[/itex] and [itex]t_2[/itex] is a linear function of t.

    The probability that project b will be complete at time t after it begins is given as:
    [tex]
    \[ p_b(t) = \left\{ \begin{array}{ll}
    0 & \mbox{if t \leq t_3$} \\
    \frac {(t - t_3)}{t_4-t_3} & \mbox{if t_3 \leq t \leq t_4$} \\
    1 & \mbox{if t_4 \leq t}
    \right. \][/tex]


    Project b will begin the moment project a is complete. Given t, what is the probability [itex]p_{ab}(t)[/itex] that both projects will be complete at time t. Obviously,
    [tex]
    \[ p_{ab}(t) = \left\{ \begin{array}{ll}
    0 & \mbox{if t \leq t_1 + t_3$} \\
    1 & \mbox{if t_2 + t_4 \leq t}
    \right. \][/tex].

    What is [itex]p_{ab}(t)[/itex] for [itex]t_1 + t_3 \leq t \leq t_2 + t_4[/itex]?
     
  2. jcsd
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